Class 12 Maths Three Dimensional Geometry | Intercept form of the equation of a plane |

**Intercept form of the equation of a plane**

Let the equation of the plane be Ax+By+Cz+ D=0 (where D≠0). (1)

Let the plane make intercepts a, b, c on x, y and z axes, respectively.

The plane meets x, y and z-axes at(a,0,0), (0,b,0) ,(0,0,c) respectively.

Therefore putting (0,0,c) we get, A(0)+B(0)+C(c)+D =0

=>C= - (D/c).

Similarly by putting (0,b,0) we get, A(0)+B(b)+C(0)+D=0

=> B=(-D/b)

And by putting (a,0,0) we get, A(a)+B(0)+C(0)+D=0

=>A=(-D/a)

Substituting the values of A,B and C in (1) we get,

(-D/a)x +(-D/b)y+(-D/c)z +D =0

=> (-D) ((x/a) + (y/b) + (z/c) – 1) =0.

**(x/a)+(y/b) + (z/c) =1. **Hence Proved.

** **

__Problem:-__

Find the intercepts cut off by the plane 2x+y-z=5.

__Answer:-__

Considering the equation 2x+y-z=5. Dividing both sides of the equation by 5, we obtain:

(2/5) x+(y/5)-(z/5) =1

=>(x)/ (5/2) + (y/5) + (z/-5) =1 (2)

It is known that the equation of a plane in intercept form is, (x/a) +(y/b) +(c/z) =1, wherea, b, c are the intercepts cut off by the plane at x, y, z axes respectively.

Therefore a=(5/2), b =5, and c=-5.

Thus the intercepts cut off by the plane are (5/2),(5) and (-5).

__Problem:-__

Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

__Answer:-__

The equation of the plane ZOX is y = 0.

Any plane parallel to it is of the form, y = a.

Since the y-intercept of the plane is 3,

a = 3

Thus, the equation of the required plane is y = 3.

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